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I am trying to procure a generlization of the following result:

$\displaystyle \int_0^\infty \frac{e^{-cx} - e^{-dx}}{x} dx = \log(d/c)$ for $0<c<d$

This result is obtained by considering $$\int_a^b \int_c^d e^{-xy} dy dx = \int_c^d \int_a^b e^{-xy} dx dy$$

And the hint tells us to begin with

$$\int_a^b \int_c^d \varphi' (xy) dy dx$$

and introduce hypotheses in $\varphi$ where needed to justify the argument.

The problem I have is that this problem is very vague (doesn't tell us what result we should get, and doesn't tell us what "suitable hypotheses" are), and I don't see a natural generalization of the result. However, this is what I have so far:

The first hypothesis I introduce is that $\frac{d}{dy} \varphi = \frac{d}{dx} \varphi$, so that I can switch integrals.

So, I begin by integrating the inner integrals:

$$\int_a^b \varphi(dx) - \varphi(cx) dx = \int_c^d \varphi(ay) - \varphi(by) dy.$$

Let $a \to 0$ and $b \to \infty$. So I suppose that $\lim \limits_{a \to 0} \varphi(ay) = A$ and $\lim \limits_{b \to \infty} \varphi(by) = B$ for fixed, finite $y$. Now this becomes

$$\int_0^\infty \varphi(dx) - \varphi(cx) dx = \int_c^d (A - B) dy$$

but now I feel like I've gone off track because the result is different from the result that we're supposed to be generalizing.

MT_
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1 Answers1

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Begin with

$$\int_a^b \int_c^d \varphi' (xy) dy dx$$

Note that $\frac{1}{x}\frac{\partial \varphi}{\partial y} = \frac{1}{y}\frac{\partial \varphi}{\partial x}$. Thus, we have $$\begin{align} \int_a^b \int_c^d \varphi' (xy) dy dx &=\int_a^b \int_c^d \frac{1}{x}\frac{\partial \varphi (xy)}{\partial y} dy dx \\ &=\int_a^b \frac{1}{x} \left( \varphi (dx)-\varphi (cx)\right) dx \end{align}$$

We also have

$$\begin{align} \int_a^b \int_c^d \varphi' (xy) dy dx &=\int_a^b \int_c^d \frac{1}{y}\frac{\partial \varphi (xy)}{\partial x} dy dx \\ &=\int_c^d \frac{1}{y} \left( \varphi (by)-\varphi (ay)\right) dy \end{align}$$

Assume that $\lim_{z \to \infty} \varphi (z) =0$. Then, as $a \to 0$ and $b \to \infty$, we see that

$$\begin{align} \int_0^{\infty} \int_c^d \varphi' (xy) dy dx &=- \varphi (0) \int_c^d \frac{1}{y} dx \\ &=-\varphi (0) \log (d/c) \end{align}$$

Putting it all together, we have

$$\int_0^{\infty} \frac{\left( \varphi (cx)-\varphi (dx)\right)}{x} dx =\varphi (0) \log (d/c)$$

which generalizes the result!

Mark Viola
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  • Nice! This is probably exactly what the question wants from me. Is this an intuitively obvious result? Personally, given the starting point, I had no idea where it was going to take me. – MT_ Mar 26 '15 at 18:43
  • I just began looking at integrating in both orders. Then, the result just popped out. I hope that this answer is the best answer I can provide. Let me know how else I can help – Mark Viola Mar 26 '15 at 18:51
  • No, this is great, thanks! – MT_ Mar 26 '15 at 19:14
  • My pleasure! Glad to help. – Mark Viola Mar 26 '15 at 19:21